(c)2023 UW Math Dept
licensed under a Creative Commons
By-NC-SA 4.0 International License.
Consequences of the rules of probability
1. (Warm-up) We sample a point (𝑥, 𝑦) uniformly from [0, 1] × [0, 2]. Let Ω = [0, 1] × [0, 2]. Compute
the probability of the following events:
(a) 𝐴 = {(𝑥, 𝑦) ∈ Ω : 𝑥 > 𝑦}
Solution note:
1
4
(b) 𝐵 = {(𝑥, 𝑦) ∈ Ω : 𝑥 ≤ 𝑦}
Solution note:
3
4
(c) 𝐶 = {(𝑥, 𝑦) ∈ Ω : 𝑥 < 21 }
Solution note:
1
2
(d) 𝐷 = {(𝑥, 𝑦) ∈ Ω : 𝑥 > 34 }
Solution note:
1
4
2. (Quick check) Consider the same set-up as Problem 1. Use set operations to express each event
below in terms of the events from Problem 1. There may be multiple answers.
(a) The event Ω.
Solution note: We have Ω = 𝐴 ∪ 𝐵. Importantly here: 𝐴 and 𝐵 are complements of each
other, and in general, 𝐴 ∪ 𝐴𝑐 = Ω.
(b) The event that both 𝑥 > 𝑦 and 𝑥 > 43 .
Solution note: This event can be expressed as 𝐴 ∩ 𝐷.
(c) The event that
1
2
≤ 𝑥 ≤ 34 .
Solution note: This event can be expressed as 𝐶 𝑐 ∩ 𝐷 𝑐 or (𝐶 ∪ 𝐷)𝑐 . From (a), we also have
Ω = 𝐴 ∩ 𝐵 so that we also have that this event is (𝐴 ∩ 𝐵) \ (𝐶 ∪ 𝐷).
3. We flip a fair coin until we have seen at least one heads and one tails and then we stop. We had
previously modeled this experiment with sample space Ω = {2, 3, . . . , } ∪ {∞} and with probabilities P({∞}) = 0 and
P({𝑘 }) =
1
,
2𝑘−1
𝑘 = 2, 3, . . . .
We want to compute the probability of flipping the coin at least 4 times. Letting 𝐴 be this event
of interest, we do this in multiple steps:
(a) Describe the event 𝐴𝑐 and express P(𝐴) in terms of P(𝐴𝑐 ).
Solution note: The event 𝐴𝑐 is the event that we flip the coin only 2 or 3 times. Since
Ω = 𝐴 ∪ 𝐴𝑐
with 𝐴 and 𝐴𝑐 disjoint, we have
1 = P(Ω) = P(𝐴) + P(𝐴𝑐 ) =⇒ P(𝐴𝑐 ) = 1 − P(𝐴),
by countable additivity.
(b) Now decompose 𝐴𝑐 into two simpler events. Compute the probability of the two simpler
events.
Solution note: We have 𝐴𝑐 = {2, 3} = {2} ∪ {3} with P({2}) =
1
2
and P({3}) = 41 .
(c) Use your answer in (b) to compute P(𝐴𝑐 )
Solution note: We again using countable additivity:
P(𝐴𝑐 ) = P({2, 3}) = P({2}) + P({3}) =
1 1 3
+ = .
2 4 4
(d) Put your answer in (a) and (c) together to compute P(𝐴).
Solution note: Finally, we arrive at
P(𝐴) = 1 − P(𝐴𝑐 ) = 1 −
3 1
= .
4 4
(e) How do you think the probability of flipping the coin at least 6 times compares to your
answer in (d)? Explain.
Solution note: It will be smaller. The event 𝐴 can be decomposed into the union of two
disjoint events: the event 𝐵 where there are at least 6 coin flips and the event 𝐶 where there
are 4 or 5 coin flips. Since P(𝐶) must be ≥ 0, then P(𝐴) = P(𝐵 ∪ 𝐶) = P(𝐵) + P(𝐶) ≥ P(𝐵)
by countable additivity.
4. In our version of Bingo, balls numbered 1 through 25 are removed, one at a time, from a rotating
cage. After 2 balls have been removed, we want to know our chances that of seeing a square (e.g.
4) or a cube (e.g., 8). We do this in several steps.
(a) What is the chance of not seeing a square? What is the chance of seeing a square?
Solution note: If we do not keep track of the order in which balls are selected, then there
are 25×24
2 possible outcomes. Among these outcomes, there are
(25 − 5)(25 − 6)
2
outcomes in which we don’t see a square. (Note: there are 5 squares between 1 and 25.)
Thus, the chance of not seeing a square is:
20 × 19
.
25 × 24
Using this answer, the chance of seeing a square is thus:
1−
20 × 19
.
25 × 24
(b) What is the chance of not seeing a cube? What is the chance of seeing a cube?
Solution note: Using same logic from (a) and given there are 2 cubes between 1 and 25,
the chance of not seeing a cube is
(25 − 2) × (25 − 3) 23 × 22
=
.
25 × 24
25 × 24
The chance of not seeing a cube is then
1−
23 × 22
.
25 × 24
(c) What is the chance of seeing a square and a cube?
Solution note: There are 5 squares and 2 cubes. The number 1 is both a cube and a square.
Therefore, we have 24 outcomes in which 1 is selected, and, we have 4 outcomes where a
square and the cube other than 1 is selected. This yields 28 outcomes, and the chance of
seeing a square and a cube is
28
56
=
25 × 24/2 25 × 24
(d) Use your answers from (a)–(c) to compute the probability of seeing a square or a cube.
Hint. Use the inclusion-exclusion formula:
P(𝐴 ∪ 𝐵) = P(𝐴) + P(𝐵) − P(𝐴 ∩ 𝐵)
Solution note: Let 𝐴 be the event of seeing a square and 𝐵 be the event of seeing a cube.
In (c), we considered event 𝐴 ∩ 𝐵. We are interested in 𝐴 ∪ 𝐵. Using the hint, we have
P(𝐴 ∪ 𝐵) = P(𝐴) + P(𝐵) − P(𝐴 ∪ 𝐵)
20 × 19
23 × 22
56
=1−
+1−
−
25 × 24
25 × 24 25 × 24
20 × 19 + 23 × 22 + 56
=2−
75 × 74
Consequences. To summarize, we have several strategies for using use set operations to
compute the probability of events. These include:
• Complements: P(𝐴) = 1 − P(𝐴𝑐 )
Í
• Decomposition: P(𝐴) = 𝑖 P(𝐴𝑖 ) for countable disjoint 𝐴𝑖 with 𝐴 = ∪𝑖 𝐴𝑖 .
• Monotonicity: 𝐴 ⊆ 𝐵 implies P(𝐴) ≤ P(𝐵)
• Inclusion-exclusion for 2 events:
P(𝐴 ∪ 𝐵) = P(𝐴) + P(𝐵) − P(𝐴 ∩ 𝐵)
or 3 events:
P(𝐴 ∪ 𝐵 ∪ 𝐶) = P(𝐴) + P(𝐵) + P(𝐶) − P(𝐴 ∩ 𝐵) − P(𝐵 ∩ 𝐶) − P(𝐴 ∩ 𝐶) + P(𝐴 ∩ 𝐵 ∩ 𝐶)
or n events:
P(∪𝑛𝑖=1𝐴𝑖 )
=
𝑛
∑︁
𝑘=1
(−1)𝑘+1
∑︁
𝑖 1 <𝑖 2 <...<𝑖𝑘
P(𝐴𝑖 1 ∩ 𝐴𝑖 2 ∩ . . . ∩ 𝐴𝑖𝑘 ).